utkalnayak wrote:Is x > 1?
1. (x+1)(|x|-1) > 0
2. |x| < 5

Statement 1: (x+1)(|x|-1) > 0
Determine the CRITICAL POINTS: the values where (x+1)(|x|-1) = 0.
(x+1)(|x|-1) = 0 when x=-1 or x=1.
To determine the ranges where (x+1)(|x|-1) > 0, test one value to the left and right of each critical point.

x<-1:
If we plug x=-2 into (x+1)(|x|-1) > 0, we get:
(-2+1)(|-2|-1) > 0
(-1)(1) > 0
-1 > 0.
Doesn't work.
Thus, x<-1 is not a valid range.

-1<x<1:
If we plug x=0 into (x+1)(|x|-1) > 0, we get:
(0+1)(|0|-1) > 0
(1)(-1) > 0
-1 > 0.
Doesn't work.
Thus, -1<x<1 is not a valid range.

x>1:
If we plug x=2 into (x+1)(|x|-1) > 0, we get:
(2+1)(|2|-1) > 0
(3)(1) > 0
3 > 0.
This works.
Thus, x>1 is a valid range.

Since x>1 is the only valid range, (x+1)(|x|-1) > 0 implies that x>1.
SUFFICIENT.

Statement 2: |x| < 5
It's possible that x=2, in which case x>1.
It's possible that x=0, in which case x<1.
INSUFFICIENT.

The correct answer is A.

As a general rule, when I see a setup where S1 is very complicated, and S2 is so obviously not sufficient that I can evaluate it in my head in three seconds, I become very very suspicious that the more complicated statement will be sufficient on its own.

Is x > 1?
1. (x+1)(|x|-1) > 0

The algebra looks hairy here, so I'd rather pick numbers. Say x = 2. That satisfies the statement, so clearly, I can get a YES to the question. Now the question is can I get a NO? Well, if x = 1/2, the product will be negative, so that won't work. And If x = -2, the product is still negative. Test away, and you'll see that only numbers bigger than one will work here, so S1 is sufficient. (Essentially, if -1 < x < 1 the first term will be positive and the second term will be negative, yielding a negative product. And if x < -1, the first term will be negative and the second term will be positive, again yielding a negative product. So then x is going to have to be bigger than 1.)


2. |x| < 5
x = 3 or x = 0; Not sufficient; (and this statement insulted our collective intelligence.)

The one thing I would make clear is that for Statement 1 to work, both left side terms have to be either negative or positive, and neither can be 0.

So when you are plugging in, you don't need to actually solve the whole thing. You just need to see if you are creating either two positive terms or two negative terms. If you are not, then the number you are using doesn't work. If you are, then it does.

Here's another algebraic approach.

Since |x| = √x², S1 really gives us

(x + 1)(√x² - 1) > 0

√x² * (x + 1) - 1 * (x + 1) > 0

√x² * (x + 1) > (x + 1)

At this point, we know that √x² is not 0 or 1. (If it were, both sides would be equal, but we're told that the left hand side is greater.) We thus have two cases to consider.

Case 1:: (x + 1) > 0

In this case, we divide both sides of the equation to get √x² > 1. This reduces to x > 1 or -1 > x, but the second case is impossible. (It would contradict (x + 1) > 0, i.e. x > -1, which we assumed to get here in the first place.)

Case 2:: 0 > (x + 1)

In this case, we divide both sides of the equation to get √x² < 1. This reduces to -1 < x < 1, but this is also impossible, since it contradicts 0 > (x + 1), or - 1 > x. Hence Case 2 is entirely impossible.

So the ONLY possibility is x > 1, and S1 is sufficient.